I am trying to proof that

$L = \{ 0^11^2...0^{n-1}1^n0^{n-1}...1^20^1\}$ where $n >= 0$ is not a regular language.

So my method is to put

$S = 0^11^2...0^{n-1}$

$W = S1^nS^R$

And then proof $S^R$ is not a regular language using pumping lemma. But as my understanding goes, the closure property is for regular language only and not the other way around. So From above I've got that

$S$ and $S^R$ is not a regular language. But $1^n$ is a regular language. So how to proof that $W$ is not a regular language?

  • 1
    $\begingroup$ You can't write $W = S1^nS^R$, since that would allow for, say, $01100011000110$. If you write it as a concatenation of languages like this, you cannot control the value of $n$ between the two definitions. $\endgroup$ – Theo Bendit Sep 23 '18 at 8:10
  • 3
    $\begingroup$ Note: If $A=\{\,0^p\mid p\text{ prime}\,\}$ and $B=0^*$, then $AB$ is regular $\endgroup$ – Hagen von Eitzen Sep 23 '18 at 8:45
  • $\begingroup$ Why don't you apply the pumping lemma to $L$? $\endgroup$ – Hagen von Eitzen Sep 23 '18 at 9:41
  • $\begingroup$ How exactly? (Normally I work with $W = XYZ$ and X = $a^i$, $Y = a^j$ but in this case, the variable is at the centre and I don't know how to prove). It would be so kind if you could give me a hint. Thanks. Also from the first comment, I just notice that. Thank you. $\endgroup$ – Wakeme UpNow Sep 23 '18 at 10:10
  • $\begingroup$ For the title question: consider the counterexample $A = \{ 0^m 1^n \mid m \ge n \}$ and $B = 1^*$. Then $AB = 0^* 1^*$ is regular. $\endgroup$ – Daniel Schepler Sep 23 '18 at 15:38

Okay, I have asked my senior for help

So what he told is

Let $L = \{ 0^11^2...0^{n-1}1^n0^{n-1}...1^20^1\}$. Given $W = XYZ$ for $\forall W\in L$

We got $|W| = n^2$. For arbitrary Y, $1\le|Y|\le n$

Consider $n^2 \lt n^2 + 1 \le |XY^2Z| \le n^2 + n < n^2 + 2n < (n + 1)^2$

So $n^2 < |XY^2Z| < (n+1)^2$

Thus proving $XY^2Z \notin L$. From pumping lemma we can conclude that $L$ is not regular.

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